POJ1023_The Fun Number System_模拟二进制运算

The Fun Number System

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10829 Accepted: 3674

Description

In a k bit 2's complement number, where the bits are indexed from 0 to k-1, the weight of the most significant bit (i.e., in position k-1), is -2^(k-1), and the weight of a bit in any position i (0 ≤ i < k-1) is 2^i. For example, a 3 bit number 101 is -2^2 + 0 + 2^0 = -3. A negatively weighted bit is called a negabit (such as the most significant bit in a 2's complement number), and a positively weighted bit is called a posibit. 
A Fun number system is a positional binary number system, where each bit can be either a negabit, or a posibit. For example consider a 3-bit fun number system Fun3, where bits in positions 0, and 2 are posibits, and the bit in position 1 is a negabit. (110)Fun3 is evaluated as 2^2-2^1 + 0 = 3. Now you are going to have fun with the Fun number systems! You are given the description of a k-bit Fun number system Funk, and an integer N (possibly negative. You should determine the k bits of a representation of N in Funk, or report that it is not possible to represent the given N in the given Funk. For example, a representation of -1 in the Fun3 number system (defined above), is 011 (evaluated as 0 - 2^1 + 2^0), and 
representing 6 in Fun3 is impossible.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. Each test case is given in three consecutive lines. In the first line there is a positive integer k (1 ≤ k ≤ 64). In the second line of a test data there is a string of length k, composed only of letters n, and p, describing the Fun number system for that test data, where each n (p) indicates that the bit in that position is a negabit (posibit). 
The third line of each test data contains an integer N (-2^63 ≤ N < 2^63), the number to be represented in the Funk number 
system by your program.

Output

For each test data, you should print one line containing either a k-bit string representing the given number N in the Funk number system, or the word Impossible, when it is impossible to represent the given number.

Sample Input

23pnp64ppnn10

Sample Output

Impossible1110

大致题意：

我们知道正常的二进制，第k位的位权是2的k-1次方。这里给出一个奇特的计数系统。给定k个字母，每个字母为p或者n。从右往左数，第k位若是p，则位权为2的k-1次方，若是n，则位权为-2的k-1次方。问题是，给出一组pn序列，求在这样的计数系统下，能否表示给定的十进制数字n。若能，输出该表示。否则，输出Impossible。

大体思路：

整体上就是模拟二进制的表示。

先讨论n为正数的情况。

1)把给定的数字转化为2进制。

2)把二进制从低位到高位依次转化为特殊进制。转化规则如下：

若该位上的数字为0，那么结果中该位也是0。

若该位是1，且位权为p，则结果中的该位也是1。

若该位是1，且位权为n，则这个1应看做从前一位借位后减去本位上的1得到的。所以这种情况的处理方法是，结果中的该位填1，原二进制的下一位加1。

以上步骤中的二进制表示和结果可以同时在一个数组中表示出来。

若n为负数，处理方法不变。只需要把对p、n的操作互换就可以了。

解题过程：

一开始没有初始化数组，导致处理进位时混乱了。

#include<cstdio>#include<cstring>char A[70];int In[70],Ans[70];int t,k,inn;long long N;char x;bool po;int main(){//freopen("in.txt","r",stdin);int i;scanf("%d",&t);//样例数while(t--){memset(In,0,sizeof(In));//初始化数组scanf("%d",&k);//位数for(i=k-1;i>=0;i--)scanf(" %c",&A[i]);//输入p/nscanf("%I64d",&N);//输入待表示的数字if(N<0) N=-N,po=0;//判断正负else po=1;inn=0;while(N)//转化为正常的二进制In[inn++]=N%2,N/=2;if(po==1) for(i=0;i<inn;i++)//处理正数{if(In[i]>=2){//若不符合二级制产生进位动作if(i==inn-1) inn++;In[i]-=2,++In[i+1],--i;//这里的i--的意思是continue继续循环后仍处理当前位continue;}if(In[i]==0) Ans[i]=0;else if(A[i]=='p') Ans[i]=1;else if(A[i]=='n') {Ans[i]=1,++In[i+1];if(i==inn-1) inn++;}}else for(i=0;i<inn;i++)//处理负数的情况{if(In[i]>=2){if(i==inn-1) inn++;In[i]-=2,++In[i+1],--i;continue;}if(In[i]==0) Ans[i]=0;else if(A[i]=='n') Ans[i]=1;//此处的操作与正数相反else if(A[i]=='p'){Ans[i]=1,++In[i+1];if(i==inn-1) inn++;}}if(inn>k) printf("Impossible\n");else {for(i=inn;i<k;i++) Ans[i]=0;for(i=k-1;i>=0;i--)printf("%d",Ans[i]);printf("\n");}}return 0;}